First, we compute a4 using the recurrence relation:
a4=2a3−c=2(12−3c)−c=24−6c−c=24−7c.
Then, we find the sum:
∑i=14ai=a1+a2+a3+a4=3+(6−c)+(12−3c)+(24−7c)
=45−11c.
Setting this greater than or equal to 23:
45−11c≥23
Solving for c gives:
45−23≥11c⇒22≥11c⇒c≤2.
Thus, the range of values of c that satisfies the condition is:
c≤2.